91Ó°ÊÓ

Partial Fraction Decomposition is a technique used to rewrite rational expressions as a sum of simpler fractions. This is particularly useful in integration when the expression's denominator is a product of linear factors. In the given exercise, we encounter the expression \(\frac{1}{u(a+b u)}\), which can be tough to integrate directly.To simplify, we express this as two separate fractions:

• \(\frac{A}{u}\)
• \(\frac{B}{a+b u}\)

Our task is to find constants \(A\) and \(B\) that make this equivalence true. By equating the original expression with the decomposed form and clearing the denominators, we solve for these constants through a system of linear equations:

• \(Aa = 0\)
• \(Ab + B = 1\)

From the first equation, \(A = 0\). Then by substituting \(A\) in the second equation, we find \(B = \frac{1}{a}\). Understanding partial fraction decomposition is crucial as it simplifies the integration process, especially for rational functions with polynomial denominators.

In calculus, we deal with two types of integrals:

• Indefinite integrals, which represent a general form of an anti-derivative.
• Definite integrals, which calculate the net area under a curve within a specific interval.

In this exercise, we focus on finding the indefinite integral of the expression \(\int \frac{du}{u(a+b u)}\). By using partial fraction decomposition, we break down the integral into simpler terms:\[\int \left( \frac{1}{u} - \frac{1}{a+b u} \right) du\]Each of these terms is standard and recognizable from basic integral rules:

• \(\int \frac{1}{u} \, du = \ln|u| \)
• \(\int \frac{1}{a+bu} \, du = \frac{1}{b} \ln|a+bu| \)

This generates an expression of logarithms that we can combine and simplify to reach the final result. Indefinite integrals always include a constant of integration, noted as \(C\), because they represent a family of functions.

The natural logarithm, denoted as \(\ln\), is an essential part of integration when dealing with expressions involving rational functions of the form \(\frac{1}{x}\). In this exercise, the integration of terms resulted in \(\ln|u|\) and \(\frac{1}{b} \ln|a+bu|\), which are classic natural logarithm forms.When integrating expressions like \(\frac{1}{x}\), it directly translates to the natural logarithm because the derivative of \(\ln|x|\) is \(\frac{1}{x}\). The expression \[\ln|u| - \frac{1}{b} \ln|a+bu|\] can be further simplified by applying the properties of logarithms, such as combining them:\[\ln \left(\frac{|u|}{(a+bu)^{1/b}}\right)\]Understanding how logarithms work in the context of integration allows for effective simplification and transformation of integral results. This step prepares us for the final expression: \[-\frac{1}{a} \ln \frac{a+bu}{u} + C\]. Leveraging the logarithmic properties makes it easier to manipulate and understand the mathematical expressions obtained through integration.